En esta sección se propone revisar en el plano, la rotación que sucede sobre una cónica, a través de los vectores propios asociados a la matriz de los coeficientes.
Se considera entonces una ecuación de la forma
a
x
2
+
b
x
y
+
c
y
2
+
d
x
+
e
y
+
f
=
0
{\displaystyle ax^{2}+bxy+cy^{2}+dx+ey+f=0}
(3.1a
a
>
0
,
{\displaystyle a>0,}
b
≠
0
{\displaystyle b\neq 0}
(3.1b
M
=
[
a
b
2
b
2
c
]
{\displaystyle M=\left[{\begin{array}{cc}a&{\frac {b}{2}}\\{\frac {b}{2}}&c\end{array}}\right]}
(3.2a
d
e
t
M
=
1
4
(
4
a
c
−
b
2
)
{\displaystyle detM={\frac {1}{4}}(4ac-b^{2})}
(3.2b
b
2
−
4
a
c
{\displaystyle b^{2}-4ac}
discriminante de la matriz M.
Valores y Vectores Propios
editar
Los valores propios se pueden obtener de la correspondiente ecuación característica, la cual es
4
λ
2
−
4
(
a
−
c
)
λ
−
(
b
2
−
4
a
c
)
=
0
{\displaystyle 4\lambda ^{2}-4(a-c)\lambda -(b^{2}-4ac)=0}
(3.3
4
(
λ
−
a
)
2
=
4
(
c
−
a
)
λ
+
4
a
2
+
b
2
−
4
a
c
{\displaystyle 4(\lambda -a)^{2}=4(c-a)\lambda +4a^{2}+b^{2}-4ac}
(3.4
λ
1
,
2
=
1
2
[
(
a
+
c
)
±
(
a
−
c
)
2
+
b
2
]
{\displaystyle \lambda _{1,2}={\frac {1}{2}}\left[(a+c)\pm {\sqrt {(a-c)^{2}+b^{2}}}\right]}
λ
1
,
2
=
1
2
[
(
a
+
c
)
±
t
r
2
M
−
4
d
e
t
M
]
{\displaystyle \lambda _{1,2}={\frac {1}{2}}\left[(a+c)\pm {\sqrt {tr^{2}M-4detM}}\right]}
3.1b
se concluye:
NOTA 1 los valores propios
λ
1
,
λ
2
{\displaystyle \lambda _{1},\;\;\lambda _{2}}
b
2
>
0
{\displaystyle b^{2}>0}
Sea
λ
1
{\displaystyle \lambda _{1}}
λ
1
=
1
2
[
(
a
+
c
)
−
(
a
−
c
)
2
+
b
2
]
{\displaystyle \lambda _{1}={\frac {1}{2}}\left[(a+c)-{\sqrt {(a-c)^{2}+b^{2}}}\right]}
(3.5a
2
(
λ
2
−
a
)
=
−
[
(
a
−
c
)
+
(
a
−
c
)
2
+
b
2
]
{\displaystyle 2(\lambda _{2}-a)=-\left[(a-c)+{\sqrt {(a-c)^{2}+b^{2}}}\right]}
(3.5b
b
≠
0
{\displaystyle b\neq 0}
N
2
<
N
2
+
b
2
{\displaystyle N^{2}<N^{2}+b^{2}}
NOTA 2
(
a
−
c
)
+
(
a
+
c
)
2
+
b
2
>
0
{\displaystyle (a-c)+{\sqrt {(a+c)^{2}+b^{2}}}>0}
2
(
λ
1
−
a
)
<
0
{\displaystyle 2(\lambda _{1}-a)<0}
λ
1
<
a
{\displaystyle \lambda _{1}<a}
Ahora, de la relación con la traza de la matriz M, se tiene
2
(
λ
2
−
c
)
=
−
2
(
λ
1
−
a
)
{\displaystyle 2(\lambda _{2}-c)=-2(\lambda _{1}-a)}
λ
2
>
c
{\displaystyle \lambda _{2}>c}
λ
1
{\displaystyle \lambda _{1}}
editar
De la relación (2.6
los vectores propios X de la matriz M satisfacen
(
M
−
λ
I
)
X
=
O
{\displaystyle (M-\lambda I)X=O}
X
t
=
[
x
,
y
]
{\displaystyle X^{t}=[x,y]}
[
a
−
λ
b
2
b
2
c
−
λ
]
[
x
y
]
=
[
0
0
]
{\displaystyle \left[{\begin{array}{cc}a-\lambda &{\frac {b}{2}}\\{\frac {b}{2}}&c-\lambda \end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}0\\0\end{array}}\right]}
y
=
2
b
(
λ
−
a
)
x
{\displaystyle y={\frac {2}{b}}(\lambda -a)x}
b
≠
0
{\displaystyle b\neq 0}
λ
{\displaystyle \lambda }
λ
1
{\displaystyle \lambda _{1}}
b
y
=
2
(
λ
1
−
a
)
x
{\displaystyle by=2(\lambda _{1}-a)x}
λ
1
{\displaystyle \lambda _{1}}
[
x
y
]
=
1
b
[
b
x
b
y
]
=
x
b
[
b
2
(
λ
1
−
a
)
]
{\displaystyle \left[{\begin{array}{c}x\\y\end{array}}\right]={\frac {1}{b}}\left[{\begin{array}{c}bx\\by\end{array}}\right]={\frac {x}{b}}\left[{\begin{array}{c}b\\2(\lambda _{1}-a)\end{array}}\right]}
x
b
≠
0
{\displaystyle {\frac {x}{b}}\neq 0}
v
1
{\displaystyle v_{1}}
v
1
=
[
b
2
(
λ
1
−
a
)
]
{\displaystyle v_{1}=\left[{\begin{array}{c}b\\2(\lambda _{1}-a)\end{array}}\right]}
(3.6
negativa.
λ
2
{\displaystyle \lambda _{2}}
editar
Puesto que el segundo valor propio
λ
2
{\displaystyle \lambda _{2}}
λ
1
{\displaystyle \lambda _{1}}
v
2
{\displaystyle v_{2}}
λ
2
{\displaystyle \lambda _{2}}
v
1
{\displaystyle v_{1}}
90
o
{\displaystyle 90^{o}}
v
1
{\displaystyle v_{1}}
v
2
=
[
−
2
(
λ
1
−
a
)
b
]
{\displaystyle v_{2}=\left[{\begin{array}{c}-2(\lambda _{1}-a)\\b\end{array}}\right]}
(3.7
positiva . También
‖
v
1
‖
=
‖
v
2
‖
{\displaystyle \|v_{1}\|=\|v_{2}\|}
Siendo el propósito describir la rotación de la cónica, el ángulo correspondiente oscilará en el intervalo
(
−
π
2
,
π
2
)
{\displaystyle \left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)}
{
v
1
,
v
2
cuando
b
>
0
(rotacion negativa)
,
−
v
1
,
−
v
2
cuando
b
<
0
(rotacion positiva)
,
{\displaystyle {\begin{cases}v_{1},\;\;v_{2}&{\text{cuando }}b>0{\text{ (rotacion negativa)}},\\-v_{1},\;\;-v_{2}&{\text{cuando }}b<0{\text{ (rotacion positiva)}},\end{cases}}}
(3.8
v
1
{\displaystyle v_{1}}
v
2
{\displaystyle v_{2}}
Dados los vectores
A
→
,
B
→
{\displaystyle {\vec {A}},\;\;{\vec {B}}}
A
→
⋅
B
→
=
‖
A
→
‖
‖
B
→
‖
cos
θ
{\displaystyle {\vec {A}}\cdot {\vec {B}}=\|{\vec {A}}\|\|{\vec {B}}\|\cos \theta }
θ
∈
[
0
,
π
]
{\displaystyle \theta \in [0,\pi ]}
A
→
=
⟨
1
,
0
⟩
{\displaystyle {\vec {A}}=\langle 1,0\rangle }
B
→
=
v
1
{\displaystyle {\vec {B}}=v_{1}}
b
=
‖
v
1
‖
cos
θ
{\displaystyle b=\|v_{1}\|\cos \theta }
ϕ
∈
(
−
π
2
,
π
2
)
{\displaystyle \phi \in \left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)}
{
ϕ
=
−
arccos
(
b
‖
v
1
‖
)
,
b
>
0
ϕ
=
π
−
arccos
(
b
‖
v
1
‖
)
,
b
<
0
{\displaystyle {\begin{cases}\phi =-\arccos \left({\frac {b}{\|v_{1}\|}}\right),&b>0\\\phi =\pi -\arccos \left({\frac {b}{\|v_{1}\|}}\right),&b<0\end{cases}}}
(3.9
b
<
0
{\displaystyle b<0}
Antes de proceder a realizar un cambio de coordenadas, se revisan las opciones de signos de los valores propios.
Sobre los signos de los valores propios de la matria M
editar
Se plantea la pregunta:
¿Es posible que
λ
1
<
0
,
λ
2
<
0
{\displaystyle \lambda _{1}<0,\;\;\lambda _{2}<0}
Supóngase que ambos valores propios son negativas; en tal caso
t
r
M
<
0
{\displaystyle trM<0}
a
+
c
<
0
{\displaystyle a+c<0}
a
>
0
{\displaystyle a>0}
c
<
0
{\displaystyle c<0}
d
e
t
M
=
a
c
−
b
2
4
<
0
{\displaystyle detM=ac-{\frac {b^{2}}{4}}<0}
NOTA 3 Sólo el valor propio
λ
1
{\displaystyle \lambda _{1}}
Ahora, ¿es posible que el valor propio
λ
2
{\displaystyle \lambda _{2}}
d
e
t
M
=
0
{\displaystyle detM=0}
a
c
=
b
2
4
{\displaystyle ac={\tfrac {b^{2}}{4}}}
λ
2
=
1
2
[
(
a
+
c
)
+
(
a
−
c
)
2
+
b
2
]
{\displaystyle \lambda _{2}={\frac {1}{2}}\left[(a+c)+{\sqrt {(a-c)^{2}+b^{2}}}\right]}
a
+
c
<
0
{\displaystyle a+c<0}
a
<
0
{\displaystyle a<0}
3.1b
. Así, se tiene
NOTA 4 El mayor valor propio
λ
2
{\displaystyle \lambda _{2}}
De la relación de los valores propios con la traza se puede decir:
NOTA 5 Si
t
r
M
<
0
{\displaystyle trM<0}
λ
1
<
0
{\displaystyle \lambda _{1}<0}
0
<
λ
2
<
−
λ
1
{\displaystyle 0<\lambda _{2}<-\lambda _{1}}
Teniendo en cuenta que el valor propio
λ
1
{\displaystyle \lambda _{1}}
NOTA 6 Si
λ
1
>
0
{\displaystyle \lambda _{1}>0}
t
r
M
>
0
{\displaystyle trM>0}
d
e
t
M
>
0
{\displaystyle detM>0}
La determinación de los vectores propios
v
1
,
v
2
{\displaystyle v_{1},\;\;v_{2}}
3.1a
en función de otras variables, con el fin de facilitar la identificación de la cónica que está describiendo. Así, se proponen las variables
u
,
v
{\displaystyle u,\;v}
[
x
y
]
=
P
[
u
v
]
{\displaystyle \left[{\begin{array}{c}x\\y\end{array}}\right]=P\left[{\begin{array}{c}u\\v\end{array}}\right]}
(3.10a
P
=
[
1
‖
v
1
‖
v
1
1
‖
v
2
‖
v
2
]
{\displaystyle P=\left[{\begin{array}{lr}{\frac {1}{\|v_{1}\|}}v_{1}&{\frac {1}{\|v_{2}\|}}v_{2}\end{array}}\right]}
3.6
y (3.8
se tiene
‖
v
1
‖
=
‖
v
2
‖
{\displaystyle \|v_{1}\|=\|v_{2}\|}
P
=
1
‖
v
1
‖
[
v
1
v
2
]
{\displaystyle P={\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}
Matriz de rotación negativa
para el caso
b
>
0
{\displaystyle b>0}
P
1
=
1
‖
v
1
‖
[
v
1
v
2
]
{\displaystyle P_{1}={\frac {1}{\|v_{1}\|\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}}}
(3.10b
3.6
y (3.7
.
Matriz de rotación positiva
Para el caso
b
<
0
{\displaystyle b<0}
3.8
P
2
=
−
1
‖
v
1
‖
[
v
1
v
2
]
{\displaystyle P_{2}=-{\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}
(3.10c
P
t
M
P
=
[
λ
1
0
0
λ
2
]
{\displaystyle P^{t}MP=\left[{\begin{array}{cc}\lambda _{1}&0\\0&\lambda _{2}\end{array}}\right]}
[
x
y
]
=
P
[
u
v
]
{\displaystyle \left[{\begin{array}{c}x\\y\end{array}}\right]=P\left[{\begin{array}{c}u\\v\end{array}}\right]}
[
x
y
]
M
[
x
y
]
+
[
d
e
]
[
x
y
]
+
f
=
0
{\displaystyle \left[{\begin{array}{lr}x&y\end{array}}\right]M\left[{\begin{array}{c}x\\y\end{array}}\right]+\left[{\begin{array}{lr}d&e\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]+f=0}
[
u
v
]
[
λ
1
0
0
λ
2
]
[
u
v
]
+
[
d
e
]
P
[
u
v
]
+
f
=
0
{\displaystyle \left[{\begin{array}{lr}u&v\end{array}}\right]\left[{\begin{array}{lr}\lambda _{1}&0\\0&\lambda _{2}\end{array}}\right]\left[{\begin{array}{c}u\\v\end{array}}\right]+\left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]+f=0}
λ
1
u
2
+
λ
2
v
2
+
[
d
e
]
P
[
u
v
]
+
f
=
0
{\displaystyle \lambda _{1}u^{2}+\lambda _{2}v^{2}+\left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]+f=0}
(3.11
[
d
e
]
P
[
u
v
]
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]}
b
>
0
{\displaystyle b>0}
editar
[
d
e
]
P
1
[
u
v
]
=
1
‖
v
1
‖
[
d
e
]
[
v
1
v
2
]
[
u
v
]
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{1}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}d&e\end{array}}\right]\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]\left[{\begin{array}{c}u\\v\end{array}}\right]}
[
d
e
]
P
1
[
u
v
]
=
1
‖
v
1
‖
{
(
[
d
e
]
v
1
)
u
+
(
[
d
e
]
v
2
)
v
}
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{1}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {1}{\|v_{1}\|}}\left\{\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)u+\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)v\right\}}
(3.12
3.11
λ
1
u
2
+
[
d
e
]
v
1
‖
v
1
‖
u
+
λ
2
v
2
+
[
d
e
]
v
2
‖
v
1
‖
v
+
f
=
0
{\displaystyle \lambda _{1}u^{2}+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}v^{2}+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{\|v_{1}\|}}v+f=0}
(3.13
Se procede a completar cuadrados:
CASO 1
d
e
t
M
≠
0
{\displaystyle detM\neq 0}
Siendo cada valor propio distindo de 0, la expresión (3.13
queda
λ
1
(
u
+
[
d
e
]
v
1
2
λ
2
‖
v
1
‖
)
2
+
λ
2
(
v
+
[
d
e
]
v
1
2
λ
2
‖
v
2
‖
)
2
=
T
1
−
f
{\displaystyle \lambda _{1}\left(u+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}+\lambda _{2}\left(v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{2}\|}}\right)^{2}=T_{1}-f}
(3.14a
T
1
=
1
4
‖
v
1
‖
2
[
(
[
d
e
]
v
1
)
2
λ
1
+
(
[
d
e
]
)
2
λ
2
]
{\displaystyle T_{1}={\frac {1}{4\|v_{1}\|^{2}}}\left[{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)^{2}}{\lambda _{1}}}+{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]\right)^{2}}{\lambda 2}}\right]}
(3.14b
3.14a
representa:
{
Una elipse,
s
i
λ
1
>
0
y
T
1
>
f
,
Una hipérbola,
s
i
λ
1
<
0
y
T
1
≠
f
,
Dos rectas coincidentes,
s
i
λ
1
<
0
y
T
1
=
f
,
Un punto,
s
i
λ
1
>
0
y
T
1
=
f
,
Ningún punto,
s
i
λ
1
>
0
y
T
1
<
f
.
{\displaystyle {\begin{cases}{\text{Una elipse,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}>f,\\{\text{Una hipérbola,}}&si\;\;\lambda _{1}<0\;\;y\;\;T_{1}\neq f,\\{\text{Dos rectas coincidentes,}}&si\;\;\lambda _{1}<0\;\;y\;\;T_{1}=f,\\{\text{Un punto,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}=f,\\{\text{Ningún punto,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}<f.\end{cases}}}
(
h
,
k
)
=
(
−
[
d
e
]
v
1
2
λ
1
‖
v
1
‖
,
−
[
d
e
]
v
1
2
λ
2
‖
v
1
‖
)
{\displaystyle (h,k)=\left({\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}},{\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)}
(3.14c
v
+
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
=
±
−
λ
1
λ
2
(
u
+
[
d
e
]
v
1
2
λ
1
‖
v
1
‖
)
{\displaystyle v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}=\pm {\sqrt {\frac {-\lambda _{1}}{\lambda _{2}}}}\left(u+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)}
(3.15
CASO 2
d
e
t
M
=
0
{\displaystyle detM=0}
Por la NOTA 4, se presenta como única opción:
λ
1
=
0
(
λ
2
>
0
)
{\displaystyle \lambda _{1}=0\;\;(\lambda _{2}>0)}
3.13
queda
[
d
e
]
v
1
‖
v
1
‖
u
+
λ
2
(
v
+
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
)
2
=
T
2
−
f
{\displaystyle {\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}\left(v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{2}-f}
(3.16a
T
2
=
(
[
d
e
]
v
2
)
2
4
λ
2
‖
v
1
‖
2
{\displaystyle T_{2}={\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)^{2}}{4\lambda _{2}\|v_{1}\|^{2}}}}
(3.16b
Una parábola, si
[
d
e
]
v
1
≠
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\neq 0}
con eje de simetría paralelo al eje y y vértice en el punto
(
‖
v
1
‖
[
d
e
]
v
1
(
T
2
−
f
)
,
−
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
)
{\displaystyle \left({\frac {\|v_{1}\|}{\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}}(T_{2}-f),{\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)}
(3.16c
Dos rectas paralelas, si
[
d
e
]
v
1
=
0
y
T
2
−
f
>
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f>0}
con ecuaciones
v
=
−
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
±
1
λ
2
(
T
2
−
f
)
{\displaystyle v={\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\pm {\sqrt {{\frac {1}{\lambda _{2}}}(T_{2}-f)}}}
(3.16d
3.16a
puede representar
Una recta, si
[
d
e
]
v
1
=
0
y
T
2
−
f
=
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f=0}
Por último, puede representar
Ningún punto, si
[
d
e
]
v
1
=
0
y
T
2
−
f
<
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f<0}
editar
Ya que la matriz a considerar
P
2
{\displaystyle P_{2}}
P
1
{\displaystyle P_{1}}
[
d
e
]
P
2
[
u
v
]
=
−
1
‖
v
1
‖
{
(
[
d
e
]
v
1
)
u
+
(
[
d
e
]
v
2
)
v
}
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{2}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {-1}{\|v_{1}\|}}\left\{\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)u+\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)v\right\}}
(3.17
3.11
produce una ecuación ligeramente dintinta de (3.12
λ
1
u
2
−
[
d
e
]
v
1
‖
v
1
‖
u
+
λ
2
v
2
−
[
d
e
]
v
2
‖
v
1
‖
v
+
f
=
0
{\displaystyle \lambda _{1}u^{2}-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}v^{2}-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{\|v_{1}\|}}v+f=0}
(3.18
lo cual indica que se obtienen los resultados de la sección 3.2.3, con una diferencia:
las coordenadas del centro en el caso de la elipse y de la hipérbola, así como en el vértice para el caso de la parábola son los opuestos aditivos.
CASO 1
d
e
t
M
≠
0
{\displaystyle detM\neq 0}
Siendo cada valor propio distinto de 0, la expresión (3.18
queda
λ
1
(
u
−
[
d
e
]
v
1
2
λ
1
‖
v
1
‖
)
2
+
λ
2
(
v
−
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
)
2
=
T
1
−
f
{\displaystyle \lambda _{1}\left(u-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)^{2}+\lambda _{2}\left(v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{1}-f}
(3.19a
T
1
=
1
4
‖
v
1
‖
2
[
(
[
d
e
]
v
1
)
2
λ
1
+
(
[
d
e
]
v
2
)
2
λ
2
]
{\displaystyle T_{1}={\frac {1}{4\|v_{1}\|^{2}}}\left[{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)^{2}}{\lambda _{1}}}+{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)^{2}}{\lambda _{2}}}\right]}
(3.19b
3.14b
3.19a
representa:
{
Una elipse,
s
i
λ
1
>
0
y
T
1
>
f
,
Una hipérbola,
s
i
λ
1
<
0
y
T
1
≠
f
,
Dos rectas coincidentes,
s
i
λ
1
<
0
y
T
1
=
f
,
Un punto,
s
i
λ
1
>
0
y
T
1
=
f
,
Ningún punto,
s
i
λ
1
>
0
y
T
1
<
f
.
{\displaystyle {\begin{cases}{\text{Una elipse,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}>f,\\{\text{Una hipérbola,}}&si\;\;\lambda _{1}<0\;\;y\;\;T_{1}\neq f,\\{\text{Dos rectas coincidentes,}}&si\;\;\lambda _{1}<0\;\;y\;\;T_{1}=f,\\{\text{Un punto,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}=f,\\{\text{Ningún punto,}}&si\;\;\lambda _{1}>0\;\;y\;\;T_{1}<f.\end{cases}}}
(
h
,
k
)
=
(
[
d
e
]
v
1
2
λ
1
‖
v
1
‖
,
[
d
e
]
v
1
2
λ
2
‖
v
1
‖
)
{\displaystyle (h,k)=\left({\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}},{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)}
(3.19c
v
−
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
=
±
−
λ
1
λ
2
(
u
−
[
d
e
]
v
1
2
λ
1
‖
v
1
‖
)
{\displaystyle v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}=\pm {\sqrt {\frac {-\lambda _{1}}{\lambda _{2}}}}\left(u-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)}
(3.20
CASO 2
d
e
t
M
=
0
{\displaystyle detM=0}
Nuevamente, como en el baso b>0, por la Nota 4, se tiene la única opción:
λ
1
=
0
(
λ
2
>
0
)
{\displaystyle \lambda _{1}=0\;(\lambda _{2}>0)}
3.18
queda
−
[
d
e
]
v
1
‖
v
1
‖
u
+
λ
2
(
v
−
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
)
2
=
T
2
−
f
{\displaystyle {\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}\left(v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{2}-f}
(3.21a
T
2
{\displaystyle T_{2}}
3.16b
, la cual representa
Una parábola, si
[
d
e
]
v
1
≠
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\neq 0}
con eje de simetría paralelo al eje u y vértice en el punto
(
−
‖
v
1
‖
[
d
e
]
v
1
(
T
2
−
f
)
,
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
)
{\displaystyle \left({\frac {-\|v_{1}\|}{\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}}(T_{2}-f),{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)}
(3.21b
Dos rectas paralelas, si
[
d
e
]
v
1
=
0
y
T
2
−
f
>
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f>0}
con ecuaciones
v
=
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
±
1
λ
2
(
T
2
−
f
)
{\displaystyle v={\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\pm {\sqrt {{\frac {1}{\lambda _{2}}}(T_{2}-f)}}}
(3.21c
3.21a
puede representar
Una recta, si
[
d
e
]
v
1
=
0
y
T
2
−
f
=
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f=0}
con ecuación
v
=
[
d
e
]
v
2
2
λ
2
‖
v
1
‖
±
1
λ
2
(
T
2
−
f
)
{\displaystyle v={\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\pm {\sqrt {{\frac {1}{\lambda _{2}}}(T_{2}-f)}}}
(3.21d
3.21a
represente
Ningún punto, si
[
d
e
]
v
1
=
0
y
T
2
−
f
<
0
{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f<0}
![{\displaystyle M=\left[{\begin{array}{cc}a&{\frac {b}{2}}\\{\frac {b}{2}}&c\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0c1d4fd2192d2b7054f2afb63b89cbe7d880729)
![{\displaystyle \lambda _{1,2}={\frac {1}{2}}\left[(a+c)\pm {\sqrt {(a-c)^{2}+b^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2bb366ef11215f690f7d85dd009cc1472f1d3fd)
![{\displaystyle \lambda _{1,2}={\frac {1}{2}}\left[(a+c)\pm {\sqrt {tr^{2}M-4detM}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/920b970d55933c60e84d5b0837451188dc08af2e)
![{\displaystyle \lambda _{1}={\frac {1}{2}}\left[(a+c)-{\sqrt {(a-c)^{2}+b^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22469080d1ddaf5cb9d2abe24ba677c9759a6707)
![{\displaystyle 2(\lambda _{2}-a)=-\left[(a-c)+{\sqrt {(a-c)^{2}+b^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8beb4ba2581a39714e8a5881fc9ced2f6949c20f)
![{\displaystyle X^{t}=[x,y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca19e9586d998ab007cd922b781e8514903c5db0)
![{\displaystyle \left[{\begin{array}{cc}a-\lambda &{\frac {b}{2}}\\{\frac {b}{2}}&c-\lambda \end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}0\\0\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bba6b86cd6a435d5dd325d9aa92636c0d79bf6a9)
![{\displaystyle \left[{\begin{array}{c}x\\y\end{array}}\right]={\frac {1}{b}}\left[{\begin{array}{c}bx\\by\end{array}}\right]={\frac {x}{b}}\left[{\begin{array}{c}b\\2(\lambda _{1}-a)\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3de83eb3ce2881d92d56f55e9815985421e8b4c)
![{\displaystyle v_{1}=\left[{\begin{array}{c}b\\2(\lambda _{1}-a)\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd8d3edb8b664f30151f0610a7ae1e332cb64b9b)
![{\displaystyle v_{2}=\left[{\begin{array}{c}-2(\lambda _{1}-a)\\b\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e97053d0f5d8ca910e82f95ba9069a3237dd0295)
![{\displaystyle \theta \in [0,\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c833964ea08aa30df8b6f56664461a5499b38144)
![{\displaystyle \lambda _{2}={\frac {1}{2}}\left[(a+c)+{\sqrt {(a-c)^{2}+b^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d138626b531ff80dc64969031f29953a7efca11b)
![{\displaystyle \left[{\begin{array}{c}x\\y\end{array}}\right]=P\left[{\begin{array}{c}u\\v\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba7889d390392f5b21891b7f317c63e28e802b45)
![{\displaystyle P=\left[{\begin{array}{lr}{\frac {1}{\|v_{1}\|}}v_{1}&{\frac {1}{\|v_{2}\|}}v_{2}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2573c37aa7972ccd41e5c87d19108b1df71ac01d)
![{\displaystyle P={\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67f548c910787d2e3c5de12e6a95dbc4ccdc316b)
![{\displaystyle P_{1}={\frac {1}{\|v_{1}\|\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c70fa9bb8e09d7d1bb9c363cfa5f4c529e004485)
![{\displaystyle P_{2}=-{\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78d638a903df7c1672f6b9f9d648447248f2d54d)
![{\displaystyle P^{t}MP=\left[{\begin{array}{cc}\lambda _{1}&0\\0&\lambda _{2}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a03e5dda3fcdafb5b21b1630a249fdccb54db51f)
![{\displaystyle \left[{\begin{array}{lr}x&y\end{array}}\right]M\left[{\begin{array}{c}x\\y\end{array}}\right]+\left[{\begin{array}{lr}d&e\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]+f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1990012a9689af4cbe6c44b79316c6783877a79)
![{\displaystyle \left[{\begin{array}{lr}u&v\end{array}}\right]\left[{\begin{array}{lr}\lambda _{1}&0\\0&\lambda _{2}\end{array}}\right]\left[{\begin{array}{c}u\\v\end{array}}\right]+\left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]+f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6eecd0becbfa31a849675d18aa894ce25532bfa1)
![{\displaystyle \lambda _{1}u^{2}+\lambda _{2}v^{2}+\left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]+f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddbebe5a3d0fc805339b8d1ddee201ca063bcadc)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P\left[{\begin{array}{c}u\\v\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a5251fcfe2f039700c89ca8850f3ceeec149ef7)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{1}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {1}{\|v_{1}\|}}\left[{\begin{array}{lr}d&e\end{array}}\right]\left[{\begin{array}{lr}v_{1}&v_{2}\end{array}}\right]\left[{\begin{array}{c}u\\v\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ee55ddd2c302a0b37340a30d5e1d7fca52788c5)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{1}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {1}{\|v_{1}\|}}\left\{\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)u+\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)v\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f52906da91b5d5cd6b9f6fc8d5ae232d52e0d326)
![{\displaystyle \lambda _{1}u^{2}+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}v^{2}+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{\|v_{1}\|}}v+f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e8983895b9992ed4fe30b882c1c87062089b6f7)
![{\displaystyle \lambda _{1}\left(u+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}+\lambda _{2}\left(v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{2}\|}}\right)^{2}=T_{1}-f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/995e68af9efed7b4f9057909251fc6c0d13ad7ba)
![{\displaystyle T_{1}={\frac {1}{4\|v_{1}\|^{2}}}\left[{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)^{2}}{\lambda _{1}}}+{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]\right)^{2}}{\lambda 2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7be650e41f6cff354d34f3241756ff2cc88579d5)
![{\displaystyle (h,k)=\left({\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}},{\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71b5a86709b1a7c74688d0723fac25ebddb260a2)
![{\displaystyle v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}=\pm {\sqrt {\frac {-\lambda _{1}}{\lambda _{2}}}}\left(u+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b41dc099d8c18b1bda8409b68378c45f68c599bf)
![{\displaystyle {\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}\left(v+{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{2}-f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33efc6c3fe11fa47fccd71b56c1df782a0b4d501)
![{\displaystyle T_{2}={\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)^{2}}{4\lambda _{2}\|v_{1}\|^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82da6d79764003898daeaa97d58bb02257641347)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaa75f780298fcefb800a781757334103f55bef2)
![{\displaystyle \left({\frac {\|v_{1}\|}{\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}}(T_{2}-f),{\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/201bd166a9913d541aab3e75a0443478990d2bd3)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f848719faefe04492704acd9066f1a77179b31c0)
![{\displaystyle v={\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\pm {\sqrt {{\frac {1}{\lambda _{2}}}(T_{2}-f)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a185f52ba28e2d2c9007fd3df57735f2eda757c)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a069d89178d20d28ba1f58ecb3a9a87c740593af)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}=0\;\;y\;\;T_{2}-f<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c892b609626926ccc9bb03dec6ef23af3b740d12)
![{\displaystyle \left[{\begin{array}{lr}d&e\end{array}}\right]P_{2}\left[{\begin{array}{c}u\\v\end{array}}\right]={\frac {-1}{\|v_{1}\|}}\left\{\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)u+\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)v\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe32dbc8c286ef377745f69d75e5702a3a31c8d6)
![{\displaystyle \lambda _{1}u^{2}-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}v^{2}-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{\|v_{1}\|}}v+f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/867c364bed3c8cb058187e447092c3fe615d67b7)
![{\displaystyle \lambda _{1}\left(u-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)^{2}+\lambda _{2}\left(v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{1}-f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b49c04a69f86f0f1efc4f8e01a4fdb258fa2d93d)
![{\displaystyle T_{1}={\frac {1}{4\|v_{1}\|^{2}}}\left[{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}\right)^{2}}{\lambda _{1}}}+{\frac {\left(\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}\right)^{2}}{\lambda _{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9182b1bc46bef9c329055f4470a2b791ab20d40c)
![{\displaystyle (h,k)=\left({\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}},{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{2}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9049b492d408357ef814c1e538dfddc98f5d7a5)
![{\displaystyle v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}=\pm {\sqrt {\frac {-\lambda _{1}}{\lambda _{2}}}}\left(u-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{2\lambda _{1}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7192ae658f9abc6dbddff0290834d5f847c08afe)
![{\displaystyle {\frac {-\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}{\|v_{1}\|}}u+\lambda _{2}\left(v-{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)^{2}=T_{2}-f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63c40aaf7592b6cb5c7119f275bfae0e1d5b9efc)
![{\displaystyle \left({\frac {-\|v_{1}\|}{\left[{\begin{array}{lr}d&e\end{array}}\right]v_{1}}}(T_{2}-f),{\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b24e599d862e5b871e2658b1038e86c5f9b3fae9)
![{\displaystyle v={\frac {\left[{\begin{array}{lr}d&e\end{array}}\right]v_{2}}{2\lambda _{2}\|v_{1}\|}}\pm {\sqrt {{\frac {1}{\lambda _{2}}}(T_{2}-f)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dd2bcd1b2b87e4886b1e9e18726f89ee39e3427)